F4tbecterkidn
Answered

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250 mL of argon gas is held in a flexible vessel shown above. If
the pressure changes to 12.0 atm, what is the new volume of
the gas at 20°C?

Sagot :

Eduard22slyidn

The new volume of the gas at 20 °C is 205 mL

From the question given above, the following data were obtained:

  • Initial volume (V₁) = 250 mL
  • Initial pressure (P₁) = 10 atm
  • Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
  • New pressure (P₂) = 12 atm
  • New temperature (T₂) = 20 °C = 20 + 273 = 293 K
  • New volume (V₂) =?

With the application of the combine gas equation, we can obtain the new volume of the gas as illustrated below:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\ \\ \frac{10 \times 250}{298} = \frac{12 \times V_2}{293} \\ \\ \frac{2500}{298} = \frac{12 \times V_2}{293} \\ \\ cross \: multiply \\ \\ 298 \times 12 \times V_2 = 2500 \times 293 \\ \\ divide \: both \: side \: by \: 298 \times 12 \\ \\ V_2 = \frac{2500 \times 293}{298 \times 12} \\ \\ V_2 = 205 \: mL \\ [/tex]

Therefore, the new volume of the gas is 205 mL

Learn more about gas laws: https://brainly.com/question/6844441

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Onyebuchinnajiidn

The new volume of the gas at the given pressure and temperature is 205 mL.

The given parameters:

  • Initial volume of argon, V₁ = 250 mL
  • Initial pressure of gas, P₁ = 10 atm
  • Final pressure of the gas, P₂ = 12 atm
  • Initial temperature of the gas, T₁ = 25 ⁰C = 273 + 25 = 298 K
  • Final temperature of the gas, T₂ = 20 ⁰C = 273 + 20 = 293 K

The new volume of the gas is calculated by applying general gas equation as follows;

[tex]\frac{V_1 P_1}{T_1} = \frac{V_2P_2}{T_2} \\\\V_2 = \frac{V_1P_1 T_2}{T_1P_2} \\\\V_2 = \frac{(250\ mL) \times 10 \times 293}{298 \times 12} \\\\V_2 = 204.84 \ mL\\\\v_2 = 205 \ mL[/tex]

Thus, the new volume of the gas at the given pressure and temperature is 205 mL.

Learn more about general gas equation here:

View image Onyebuchinnaji
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